Animated Solution for Mathematics - Conic Sections: Let C be the largest circle centred at $(2,0)$ and inscribed in the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$. If $(1,\alpha )$ lies on C, then $10{\alpha }^2$ is equal to _____.

Imagine you are standing on the coordinate plane, looking at an ellipse defined by:

You have a center point at $(2,0)$, and you want to expand a circle from this center until it perfectly kisses the boundary of the ellipse. This is not just a drawing exercise; it is a profound interaction between two conic sections.

The key to unlocking this problem lies in the concept of the normal. When the circle is at its maximum size, it is tangent to the ellipse. At this point of tangency, the normal to the ellipse—the line perpendicular to the tangent—must pass through the center of the circle.

Instead of drowning in the Cartesian equation, we embrace the elegance of parametric coordinates. For an ellipse with semi-major axis $a=6$ and semi-minor axis $b=4$, any point $P$ on the curve can be represented as $(6\cos \theta ,4\sin \theta )$.

Now, we invoke the standard equation for the normal at a point $(x_1,y_1)$ on an ellipse:

Substituting our point $P(6\cos \theta ,4\sin \theta )$ and the values $a^2=36$ and $b^2=16$, we get:

Simplifying this, we arrive at $6x\sec \theta -4y\csc \theta =20$. Dividing by two, we get the beautiful, clean equation:

We know this normal must pass through our circle's center, $(2,0)$. By substituting $x=2$ and $y=0$ into our normal equation, the $y$ term vanishes, leaving us with:

This simplifies to $\sec \theta =\frac{5}{3}$, which means $\cos \theta =\frac{3}{5}$. Consequently, $\sin \theta =\frac{4}{5}$.

With these values, we find the exact coordinates of our point of tangency $P$:

The radius $r$ is simply the distance between the center $C(2,0)$ and the point $P(\frac{18}{5},\frac{16}{5})$. Using the distance formula:

The equation of our circle is $(x-2{)}^2+y^2=\frac{64}{5}$. We are told the point $(1,\alpha )$ lies on this circle.

Substituting these coordinates, we get $(1-2{)}^2+{\alpha }^2=\frac{64}{5}$, which simplifies to $1+{\alpha }^2=\frac{64}{5}$. Thus, ${\alpha }^2=\frac{64}{5}-1=\frac{59}{5}$.

The question asks for $10{\alpha }^2$. Multiplying $\frac{59}{5}$ by $10$ gives us $2\times 59=118$.

The final answer is 118.